# Cyclic Averages of Regular Polygons and Platonic Solids

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1 Cyclic Averages of Regular Polygos ad Platoic Solids Mamuka Meskhishvili Departmet of Mathematics, Georgia-America High School, 8 Chkodideli Str., Tbilisi 080, Georgia mathmamuka@gmail.com Preprit submitted to RGN Publicatios o 03/06/00 Abstract For a regular polygo P ad Platoic solid T are itroduced the cocept of the cyclic averages. It is show the cyclic averages of equal powers are the same for various P T, but their umber is characteristic of P T. Give the defiitio of a circle sphere by the vertices of P T ad o the base of the cyclic averages are established the commo metrical relatios of P T. MSC. 5M04, 4G05 Keywords ad phrases. Regular polygo, Platoic solid, circle, sphere, locus, sum of like powers, ratioal distaces problem Article type: Research article Itroductio Cosider a fiite set of poits i the plae space, the locus of poits such that the sum of the squares of distaces to the give poits is costat, is a circle sphere, whose ceter is at the cetroid of the give poits [, ]. Deote by Md, d,..., d, L a arbitrary poit i the plae space of a regular polygo Platoic solid of distaces d, d,..., d to the vertices A, A,..., A, the: d i = R + L, where R is the radius of circumscribed circle sphere of the regular polygo Platoic solid ad L is the distace betwee the poit M ad the cetroid O. For a equilateral triagle ad a arbitrary poit Md, d, d 3, L i the plae of the triagle the symmetric equatio exists 3d 4 + d 4 + d a 4 = d + d + d 3 + a,

3 It is clear, the sum of odd power cotais radicals ad ever will be [R,L]. For establishig commo properties of the P ad T discussig average of [R,L] prefered S m = m. [R,L] is much Defiitio.. The cyclic averages S m average of the sum m [R,L]. S m [] of a regular polygo Platoic solid is the We call such averages as the cyclic averages, because as we prove for fixed R ad L the cyclic averages of equal powers of differet P T are the same if they exist: if. S m = S m, O the other had for ay give P the umber of S m as well as m [R,L] is defied uiquely, so the umber of the cyclic averages is characteristic of the regular polygo. For example for a regular 3-go exists cyclic averages: while for a regular 4-go 3 cyclic averages: They are i relatios: S 3 ad S 4 3, S 4, S4 4 ad S 6 4. S 3 = S 4 ad S 4 3 = S 4 4. To demostrate the efficiecy of cyclic averages the aalogue of the relatio will be obtaied for the square. Firstly we tur i terms of R ad the cyclic averages S 3, S4 3 : the replace with d 4 + d4 + d R 4 = d + d + d R, S 3 = S 4, S4 3 = S 4 4 ad R = a ; we get Theorem.. For a arbitrary poit Md,..., d 4, L i the plae of a square: 4d 4 + d 4 + d d a 4 = d + d + d 3 + d 4 + a, where a is the side of the square. 3

4 3 Circle as Locus of Costat [R,L] Sums Theorem 3.. For a arbitrary poit Md, d,..., d, L i the plae of regular polygo P R: i= d m i where m =,...,. [ = R + L m + First we eed to prove two lemmas. m k= m k k k R k L k R + L m k ], Lemma 3.. For arbitrary positive itegers m ad, such that m <, the followig coditio k= is satisfied, where α is a arbitrary agle. Deote The real part of T is cos m α k π = 0 T = e imα + e imα π + e imα π + + e imα π. ReT = k= cos m α k π. The formula of the sum of geometric progressio gives T = e imα + e im π + e im π + + e im π e im π = cos πm + i πm =. π imα e im = e, e im π Sice m <, e im π. So T = 0, i.e. ReT = 0, which proves Lemma 3.. Remark 3.. If m, the sum always cotais α. Lemma 3.. For arbitrary positive itegers m ad, such that m < ad for a arbitrary agle α the followig coditios are satisfied: if m is odd cos m α k π = 0; k= 4

5 if m is eve cos m α k π = k= mm m. Whe m is odd, usig the power-reductio formula for cosie cos m θ = m m k=0 m cos m kθ, k we obtai cos m α k π k= = cos m α + cos m α π + + cos m α π [ m = m m m cos mα + cosm α + + m 0 m + cos m α π m + cosm α π 0 m + m cos α π + + = m m cos m α π 0 [ m m + + m cos mα + cos m m + cosm cos α π cos α + α π ] + + cos m α π α π 0 m + cosm α+cosm α π + + cosm α π + m + m cos α + cos α π + + cos α π ]. Sice m <, from Lemma 3. it follows that each sum equals zero, which proves the first part of Lemma 3.. Whe m is eve, the power-reductio formula for cosie is cos m θ = m m m + m m k=0 5 m cos m kθ. k

6 Aalogously to the case with odd m, the sum of the secod addeda vaishes, ad sice the umber of the first addeda is, the total sum equals m m m, which proves Lemma 3.. Proof of Theorem 3.. We itroduce the ew otatios A = R + L ad B = RL. The i= π m d m i = A B cos α m + A B cos α If m =, by Lemma 3. we have i= Therefore d m i = A B cos α If m >, we have i= d m i + A B cos π m α + + A B cos π m α. π + A B cos α + + A B cos π α = A. m = A m A m B m + A m B m A m 3 B 3 3 m ± B m m S = R + L. π cos α + cos α + + cos π α cos α + cos π α + + cos π α cos 3 α + cos 3 π α + + cos 3 π α + cos m α + cos m π α + + cos m π α. Accordig to Lemma 3., all sums with the egative sig vaish ad oly the sums with the positive sig remai. 6

7 i= If m is eve d m i = A m + = If m is odd i= d m i = A m + = m A m B cos α + cos π α + + cos π α + m + B m cos m α + cos m π m α m m A m + A m k B k k k k. k + A m + k= m m m A m B cos α + cos π m k= AB m m A m k B k k k + + cos m π α α + + cos π α + cos m α + cos m π α + + cos m π α k. k Usig the floor fuctio the iteger part, the obtaied results ca be combied ito a sigle formula as follows i= which proves the theorem. d m i From Theorem 3. each sums = A m + i= m k= m A m k B k k k d m i, where m =,,..., k, k are the [R,L] sums. Begiig from the m all sums of power distaces cotai α Remark 3.. For example for P 3 the sums cotai: - cos 3α, if m = 3, 4, 5; - cos 3α ad cos 6α, if m = 6, 7, 8; - cos 3α, cos 6α ad cos 9α, if m = 9, 0,. 7

8 Geerally for m the sums i= d m i such sums is beyod the scope of this article. cotai cosie of the multiples of α. The study of Therefore for P exist a umber of [R,L] sums ad if they are costat the locus for each case is a circle: Theorem 3.. The locus of poits such that the sum of the m-th power of the distaces to the vertices of a give P R is costat is a circle, if i= whose ceter is the cetroid of the P R. Remark If - If i= i= d m i d m i d m i > R m, where m =,,...,, = R m, the locus is the cetroid of the polygo. < R m, the locus is the empty set. 4 Cyclic Averages of Regular Polygos The properties of the cyclic average are as follows: Property 4.. Each regular -go has a umber of cyclic averages S, S 4,..., S. Property 4.. For fixed R ad L, the cyclic averages of equal powers of differet regular -gos are the same: S 3 = S 4 = S 5 = S 6 =, S 4 3 = S 4 4 = S 4 5 = S 4 6 =, S 6 4 = S 6 5 = S 6 6 =, S 8 5 = S 8 6 =. Property 4.3. Ay relatios i terms of the cyclic averages S m, the circumscibed radius R ad the distace L, which are satisfied for a regular -go, are at the same time satisfied for ay regular -go, where, i.e. S m ca be replaced by S m. Elimiate L from the relatios of Theorem 3. we obtai: 8

9 Theorem 4.. For ay regular -go: where m =,...,. S m = S m + I terms of S ad S 4 : m k= Theorem 4.. For ay regular -go: where m = 3,...,. m S m = S m + k= k m k m k k k k k The first two relatios of Theorem 3. implies: Theorem 4.3. For ay regular -go: R = so L = S ± S The poits o the circumscribed circle satisfy R k S R k S m k, S 4 S k S m k, 3S S 4 3S S 4 3S = S 4, Theorem 4.4. For ay poit o the circumscribed circle of the regular -go: 4. Equilateral triagle There are cyclic averages: 3 i= d i = d 4 i. i= S 3 = 3 d + d + d 3 = R + L, I geeral case from Theorem 4., for 3 [5],. S 4 3 = 3 d4 + d 4 + d 4 3 = R + L + R L. S 4 + 3R 4 = S + R. Deote by the symbol a,b,c the area of a triagle whose sides have legths a, b, c. The solutio of the system of the cyclic averages is: 9

10 Theorem 4.5. For ay poit Md, d, d 3, L ad P 3 R d = d, d = 3R + L d ± 4 3 R,L,d, d 3 = 3R + L d 4 3 R,L,d. ad For P 3 3S S 4 = d + d + d 3 3 d 4 + d 4 + d 4 3 = 6 3 d,d,d 3, R = d + d + d 3 ± d,d,d 3, L = d + d + d d,d,d 3. For ay poit o the circumscribed circle, follows the area d,d,d 3 should be zero. Ideed for the largest distace d 3 = d + d holds. 4. Square There are 3 cyclic averages: S 4 = 4 d + d + d 3 + d 4 = R + L, S 4 4 = 4 d4 + d 4 + d d 4 4 = R + L + R L, S 6 4 = 4 d6 + d 6 + d d 6 4 = R + L 3 + 6R L R + L. From Theorems 4. ad 4. Theorem 4.6. For ay regular -go, where 4: S 6 = S S + 3R 5R 4, S 6 = S 3S 4 S. From Theorem 4.6 follows: 8d 6 + d 6 + d d d + d + d 3 + d 4 3 = 6d + d + d 3 + d 4d 4 + d 4 + d d 4 4, which is equivalet to 3d + d d 3 d 4d + d 3 d d 4d + d 4 d d 3 = 0. 0

11 So holds. d + d 3 = d + d 4 Obtaied relatio has geeralizatio for regular -go. If is eve for the diametrically opposed vertices: Theorem 4.7. For ay regular -go, with eve umber of vertices = k: d + d +k = d + d +k = = d k + d k = R + L. Theorem 4.7 simplifies the system of the cyclic averages: S R = S 4 + R, d + d 3 = d + d 4; which is aalogue to systems obtaied i [6, 7]. Moreover i terms of R ad L, we get: The solutio of which is: d + d 3 = d + d 4 = R + L, d d 3 + d d 4 = R 4 + L 4. Theorem 4.8. For ay poit Md, d, d 3, d 4, L ad P 4 R: d = d, d = R + L ± 4 R,L,d, d 3 = R + L d, d 4 = R + L 4 R,L,d. For P 4 3S S = [ ] 3d + d + d 3 + d 6 4 8d 4 + d 4 + d d 4 4 = 4 d, d,d 3 = 4 d, d 3,d 4, ad R = 4 d + d 3 ± d, d,d 3 = 4 d + d 4 ± d, d 3,d 4, L = 4 d + d 3 d, d,d 3 = 4 d + d 4 d, d 3,d 4. For ay poit o the circumscribed circle the areas d, d,d 3 ad d, d 3,d 4 should be zero. Ideed, if the poit o the mior arc A A are satisfied d + d = d 3 ad d + d 4 = d 3.

12 4.3 Regular Petago, Hexago ad Heptago There are 4, 5 ad 6 cyclic averages for the P 5, P 6 ad P 7 cases, respectively: S 5 = S 6 = S 7 = R + L, S 4 5 = S 4 6 = S 4 7 = R + L + R L, S 6 5 = S 6 6 = S 6 7 = R + L 3 + 6R L R + L, S 8 5 = S 8 6 = S 8 7 = R + L 4 + R L R + L + 6R 4 L 4, S 0 6 = S 0 7 = R + L 5 + 0R L R + L R 4 L 4 R + L, S 7 = R + L R L R + L R 4 L 4 R + L + 0R 6 L 6. These systems are simplified for the regular hexago case oly. The vertices A, A 3, A 5 ad A, A 4, A 6 form two equilateral triagles, so they satisfy two cyclic relatios for P 3. Geerally for -go if divisible by 3: Theorem 4.9. For ay regular -go, if = 3l d + d +l + d +l = = d l + d l + d 3l = 3R + L, d 4 + d 4 +l + d4 +l = = d4 l + d4 l + d4 3l = 3 R + L + R L. The Theorem 4.7 ad Theorem 4.9 simplify the system of the cyclic averages for the regular hexago: d + d 4 = d + d 5 = d 3 + d 6 = R + L, d + d 3 + d 5 = d + d 4 + d 6 = 3R + L, d 4 + d d 4 5 = d 4 + d d 4 6 = 3 R + L + R L. By usig these relatios, we get explicit expressios for distaces: Theorem 4.0. For ay poit Md, d,..., d 6, L ad P 6 R: d = d, d = R + L + d ± 4 3 R,L,d, d 3 = 3R + 3L d ± 4 3 R,L,d d 4 = R + L d, d 5 = 3R + 3L d 4 3 R,L,d d 6 = R + L + d 4 3 R,L,d.,,

13 For P 6 : ad d 3S S 4 = 3 + d + + d 6 d 4 + d4 + + d d + d 3 + d 5 d 4 + d d 4 5 = 3 = 6 3 d,d 3,d 5 = 6 3 d,d 4,d 6 R = d + d 3 + d 5 ± d,d 3,d 5 = d + d 4 + d 6 ± d,d 4,d 6, L = d + d 3 + d d,d 3,d 5 = d + d 4 + d d,d 4,d 6. For ay poit o the circumscribed circle the area d,d 3,d 5 as well as d,d 4,d 6 vaishes. Ideed if the poit o the mior are A A : d + d 3 = d 5 ad d + d 6 = d Regular Octago, Noago ad Decago There are 8, 9 ad 0 cyclic averages for the P 8, P 9 ad P 0 cases, respectively. The cyclic averages from the secod to the twelfth powers are the same as for regular heptago, so we write oly ew oes: S 4 8 = S 4 9 = S 4 0 = R + L 7 + 4R L R + L 5 + 0R 4 L 4 R + L R 6 L 6 R + L, S 6 9 = S 6 0 = R + L R L R + L R 4 L 4 R + L R 6 L 6 R + L + 70R 8 L 6, S 8 0 = R + L 9 + 7R L R + L R 4 L 4 R + L 5 All three cases = 8, 9, 0 admit further simplificatios. For P 8 Theorem 4.7 gives: + 680R 6 L 6 R + L R 8 L 8 R + L, d + d 5 = d + d 6 = d 3 + d 7 = d 4 + d 8 = R + L. The vertices A, A 3, A 5, A 7 ad A, A 4, A 6, A 8 form two squares, so they satisfy additioal cyclic relatios for P 4. Geerally, if is divisible by 4: Theorem 4.. For ay regular -go, if = 4p: d 4 + d 4 +p + d 4 +p + d 4 +3p = = d 4 p + d 4 p + d 4 3p + d 4 4p = 4 R + L + R L, d 6 + d 6 +p + d 6 +p + d 6 +3p = = d 6 p + d 6 p + d 6 3p + d 6 4p = 4 R + L 3 + 6R L R + L. 3

14 For P 9 Theorem 4.9 gives: d + d 4 + d 7 = d + d 5 + d 8 = d 3 + d 6 + d 9 = 3R + L, d 4 + d d 4 7 = d 4 + d d 4 8 = d d d 4 9 = 3 R + L + R L. For P 0, from Theorem 4.7: d + d 6 = d + d 7 = d 3 + d 8 = d 4 + d 9 = d 5 + d 0 = R + L. The vertices A, A 3, A 5, A 7, A 9 ad A, A 4, A 6, A 8, A 0 form two regular petagos, so they satisfy additioal cyclic relatios for P 5. Geerally, if is divisible by 5: Theorem 4.. For ay regular -go, if = 5t d + d +t + d +t + d +3t + d +4t = = d t + d t + d 3t + d 4t + d 5t = 5R + L, d 4 + d 4 +t + d 4 +t + d 4 +3t + d 4 +4t = = d 4 t + d 4 t + d 4 3t + d 4 4t + d 4 5t = 5 R + L + R L, d 6 + d 6 +t + d 6 +t + d 6 +3t + d 6 +4t = = d 6 t + d 6 t + d 6 3t + d 6 4t + d 6 5t = 5 R + L 3 + 6R L R + L, d 8 + d 8 +t + d 8 +t + d 8 +3t + d 8 +4t = = d 8 t + d 8 t + d 8 3t + d 8 4t + d 8 5t = 5 R + L 4 + R L R + L + 6R 4 L 4. Summarize the obtaied results, we coclude: every regular -go has a umber of cyclic averages, but if is the composite umber we have additioal relatios for the distaces, which are obtaied from the cyclic averages of the -go, where is divisible of. 5 Ratioal Distaces Problem. Solutio for = 4 Is there a poit all of whose distaces to the vertices of the uit polygo are ratioal? The problem has a log history especially for the case of square. A extesive historical review is give i [6 8]. For case of a equilateral triagle aswer is positive [9]. Accordig [0] ope problems are i followig cases = 4, 6, 8, ad 4. For = 6 oly trivial poit is kow the cetroid of the uit hexago. By Theorem 4.3 the side a of the regular -go is: a si π For the uit icositetrago = 4: si π 4 = = S ± S 3S S 4. 4 ± 3S 4 S 4 4 S 4 4. S 4 4

15 The right side is the root of the fourth degree polyomial equatio with ratioal coefficiets: 8 S 4 S x 4 4S x + = 0, thus it is the algebraic umber of degree 4. O the other had si π 4 = + 3, is the algebraic umber of degree > 4 []. So, Theorem 5.. There is ot a poit i the plae that is at ratioal distaces from the vertices of the uit regular 4-go. For positive aswers for the P 4 ad P 6 cases the ecessary coditios are the ratioalities of the equal areas: - d, d,d 3 = d, d 3,d 4, if = 4; - 3 d,d 3,d 5 = 3 d,d 4,d 6, if = 6. 6 Sphere as Locus of Costat [R,L] Sums For regular polygos with differet vertices the umber of the [R,L] sums are differet too. As we see, ulike the plae case, dual Platoic solids have the same umber of the [R,L] sums: regular tetrahedro octahedro ad cube [R,L], 4 [R,L], icosahedro ad dodecahedro [R,L], 4 [R,L], 6 [R,L], To prove these, we cosider each platoic solid separately. cetered at the origi ad use simple Cartesia coordiates. 6. Regular Tetrahedro The coordiates of the vertices T 4 R: [R,L], 4 [R,L], 6 [R,L], 8 [R,L], 0 [R,L]. A, c, ±c, ±c, A 3,4 c, ±c, c ad R = 3 c. I all cases, we cosider solids Cosider a arbitrary poit i space Md, d, d 3, d 4, L with the coordiates: x, y, z. The distace betwee M ad the cetroid O of the tetrahedro: L = x + y + z. 5

16 The, d, = x c + y c + z c = R + L + c x y z, d 3,4 = x + c + y c + z ± c = R + L + cx y ± z, d 4 i = R + L + c x y z + R + L + cx y ± z = 4R + L + 4c x + y + z + x + y + z + x y + z + x + y z = 4 R + L R L. If for T 4 R: the d 4 i > 4R 4, Theorem 6.. The locus of poits i the space such that the sum of the fourth power of the distaces to the vertices of a give regular tetrahedro is costat is a sphere whose ceter is the cetroid of the tetrahedro. Remark If - If d 4 i = 4R4 the locus is the cetroid. d 4 i < 4R4 the locus is the empty set. The sums of the distaces of the power more tha 4 cotai x, y ad z like α for the plae case, so for T 4 oly the sums of the secod ad fourth powers are [R,L] sums. 6. Octahedro ad Cube The coordiates of the vertices of the octahedro T 6 R: A, ±c, 0, 0, A 3,4 0, ±c, 0, A 5,6 0, 0, ±c ad R = c. For a arbitrary poit P d, d,..., d 6, L: d, = R + L ± Rx, d 3,4 = R + L ± Ry, d 5,6 = R + L ± Rz. 6

17 Begiig from T 6 each Platoic solid has except tetrahedro diametrically opposed vertices, so for them is satisfied Theorem 4.7. For the sums of the fourth ad sixth powers: For the cube T 8 R: ad R = 3 c. 6 d 4 i = R + L ± Rx + R + L ± Ry ± R + L ± Rz = 4 R + L R L, 6 d 6 i = 6R + L 3 + 4R + L R x + y + z = 6 R + L 3 + 4R L R + L. The distaces from the P d, d,..., d 8, L: A, c, c, c, A 3,4 ±c, ±c, c, A 5,6 ±c, c, ±c, A 7,8 c, ±c, ±c d, = R + L ± cx + y + z, d 5,6 = R + L cx + z y, d 3,4 = R + L cx + y z, d 7,8 = R + L ± cx y z. The vertices A, A 3, A 5, A 7 ad A, A 4, A 6, A 8 form two regular tetrahedros, so they satisfy the regular tetrahedro relatios. Theorem 6.. For a arbitrary poit i the space, the sum of the quadruple of the distaces to the vertices of the cube which lie o parallel faces ad are edpoits of skew face diagoals, satisfies d k = d 4 k = d k = 4R + L, d 4 k R = 4 + L R L. Remark 6.. These quadruples do ot cotai the distaces to diametrically opposed vertices. Thus, 8 d 4 i = 8 R + L R L. 8 d 6 i = R + L ± cx + y + z 3 + R + L cx + y z 3 7

18 the + R + L cx + z y 3 + R + L cy + z x 3 = 8R + L 3 + 4R + L c x + y + z + x + y z + x y + z + x y z = 8 R + L 3 + 4R L R + L. If for T 6 R ad T 8 R is satisfied i= d m i > R m, = 6, 8; Theorem 6.3. The locus of poits i the space such that the sum of the sixth fourth power of distaces to the vertices of a give octahedro cube is costat is a sphere whose ceter is the cetroid of the octahedro cube. Remark If - If d m i d m i = R m the locus is the cetroid. < R m the locus is the empty set. 6.3 Icosahedro ad Dodecahedro The coordiates of the vertices of icosahedro T R: A, 0, ±c, ±cϕ, A 3,4 0, c, ±cϕ, A 5,6 ±c, ±cϕ, 0, A 7,8 ±c, cϕ, 0, A 9,0 ±cϕ, 0, ±c, A, ±cϕ, 0, c, where ϕ is the golde ratio ϕ = + 5 ad R = c + ϕ. For a arbitrary poit P d, d,..., d, L: d, = R + L cy + zϕ, d 5,6 = R + L cx + yϕ, d 9,0 = R + L cz + xϕ, d 3,4 = R + L ± cy zϕ, d 7,8 = R + L cx yϕ, d, = R + L ± cz xϕ. 8

19 The d 4 i = d 6 i = d 4 i + 8 d 4 i d 4 i = 4R + L + 4c y + z ϕ + 4R + L + 4c x + y ϕ + 4R + L + 4c z + x ϕ = R + L + 6c + ϕ x + y + z = R + L R L. d 6 i + For the sum of the eight power 8 d 6 i = 4R + L c R + L y + z ϕ d 6 i + 4R + L c R + L x + y ϕ + 4R + L c R + L z + x ϕ = R + L 3 + 4R L R + L. d 8 i = 4R + L c R + L y + z ϕ + 64c 4 y 4 + z 4 ϕ 4 + 6y z ϕ, Because d 8 i = R + L c R + L x + y + z + ϕ + 64c 4 x 4 + y 4 + z 4 + ϕ 4 + 6x y + x z + y z ϕ, + ϕ 4 = 3ϕ ad ϕ = 5 + ϕ, d 8 i = R + L 4 + 8R L R + L R4 L 4. d 0 i = 4R + L R + L 3 c y + zϕ + y zϕ + 60R + L c 4 y + zϕ 4 + y zϕ 4 = 4R + L R + L 3 c y + zϕ + 30R + L c 4 y 4 + z 4 ϕ 4 + 6y z ϕ. d 0 i = R + L R + L 3 c x + y + z + ϕ 9

20 + 30R + L c 4 + ϕ 4 x 4 + y 4 + z 4 + 3ϕ x y + x z + y z = R + L R L R + L 3 + 6R 4 L 4 R + L. The vertices of the dodecahedro T 0 R divide ito two groups, the vertices A, A,..., A 8 which form a cube ad other vertices A 9, A 0,..., A 0. The the coordiates: ad R = 3 c. A, c, c, c, A 3,4 ±c, ±c, c, A 5,6 ±c, c, ±c, A 7,8 c, ±c, ±c, A 9,0 0, ± c ϕ, ±cϕ, A, 0, c, ϕ, ±cϕ A 3,4 ± c ϕ, ±cϕ, 0, A 5,6 c, ϕ, ±cϕ, 0 A 7,8 ± cϕ, 0, ± c, A 9,0 ± cϕ, 0, c. ϕ ϕ Cosider a arbitrary poit P d, d,..., d 0, L. For the distaces d, d,..., d 8 we use the respective distaces of the cube, ad for others: y y d 9,0 = R + L c ϕ + zϕ, d, = R + L ± c ϕ zϕ, x x d 3,4 = R + L c ϕ + yϕ, d 5,6 = R + L ± c ϕ yϕ, z z d 7,8 = R + L c ϕ + xϕ, d 9,0 = R + L ± c ϕ xϕ, 0 0 d 4 i = 8R + L R L + 0 d 4 i 9 = 8R + L R L + R + L + 6c x + y + z ϕ + ϕ = 0 R + L R L. d 6 i = 8R + L 3 + 3R L R + L + R + L 3 + 3R + L 6c x + y + z ϕ + ϕ = 0 R + L 3 + 4R L R + L. 8 d 8 i = R + L ± cx + y + z 4 + R + L cx + y z 4 + R + L cx + z y 4 + R + L ± cx y z 4 = 8R + L R L R + L R4 L 4 + 8x y + 8x z + 8y z, 0

21 the 0 d 8 i = R + L R + L L 3c c 4 x 4 + y 4 + z 4 ϕ 4 + ϕ 4 + 6x y + 6x z + 6y z = R + L R + L R L R4 7x 4 + y 4 + z 4 + 6x y + 6x z + 6y z, 0 d 8 i = 0 R + L 4 + 8R L R + L R4 L 4. Like T, maximal power for T 0 which depeds o R ad L oly is 0. Ideed, d 0 i = R + L ± cx + y + z 5 + R + L cx + y z 5 + R + L cx + z y 5 + R + L ± cx y z 5 = 8R + L R + L 3 c L + 30R L c 4 x 4 + y 4 + z 4 + x y + x z + y z, d 0 i = R + L R + L 3 c L ϕ + ϕ + 30R + L c 4 ϕ 4 + ϕ4 x 4 + y 4 + z 4 + 6x y + x z + y z, d 0 i = 0 R + L R L R + L 3 + 6R 4 L 4 R + L. If for T R ad T 0 R is satisfied i= d m i > R m, =, 0, Theorem 6.4. The locus of poits i the space such that the sum of the m-th power of distaces to the vertices of a give icosahedro dodecahedro is costat is a sphere, whe m =,, 3, 4 ad 5. The ceter of the sphere is the cetroid of the icosahedro dodecahedro.

22 Remark If = R m the locus is the cetroid. - If i= i= d m i d m i < R m the locus is the empty set. 7 Cyclic Averages of Platoic Solids Summurize the obtaied results, i terms of the cyclic averages: Theorem 7.. The cyclic averages of the Platoic solids are the followig: S [4] = S [6] = S [8] = S [] = S [0] = R + L, S 4 [4] = S 4 [6] = S 4 [8] = S 4 [] = S4 [0] = R + L R L, S 6 [6] = S 6 [8] = S 6 [] = S6 [0] = R + L 3 + 4R L R + L, S 8 [] = S8 [0] = R + L 4 + 8R L R + L R4 L 4, S 0 [] = S 0 [0] = R + L R L R + L 3 + 6R 4 L 4 R + L. Elimiate L ad R from the relatios, we obtai direct relatios amog the cyclic averages of the Platoic solids. Theorem 7.. For each Platoic solid = 4, 6, 8,, 0: S 4 [] R4 = S [] + 3 R. This result for regular simplicial ad regular polytopic distaces is obtaied i [] ad [3], respectively. Theorem 7.3. For each Platoic solid, except the tetrahedro = 6, 8,, 0: S 6 [] = S [] S [] + R 8R 4, S 6 [] = S 4 [] 3S [] S []. Theorem 7.4. For the icosahedro ad the dodecahedro =, 0: S 8 [] S [] 4 = 8R S [] R S [] + 5 R S [] R, S 0 [] S [] 5 = 8R S 5 [] S [] R 3 S [] + R S [] R S 8 [] = 5 9S 4 S 0 [] = S [] S4 [] [] + S 4 [] S 9S [] 8S []. [] 6S [] 4,,

23 Like the plae cases, i some space cases we have additioal relatios. Each Platoic solid, except the tetrahedro satisfies Theorem 4.7 ad for the cube ad the dodecahedro Theorem 6.. For the radius of the circumscribed sphere ad the distace betwee the poit ad the cetroid: Theorem 7.5. For each Platoic solid = 4, 6, 8,, 0: R = S [] ± 4S [] 3S 4 [], L = S [] 4S [] 3S 4 []. so The poits o the circumscribed sphere satisfy 4S [] = 3S 4 [], Theorem 7.6. For ay poit o the circumscribed sphere of each Platoic solid =4, 6, 8,, 0: 4 d i = 3 d 4 i. 8 Coclusio I the preset paper, we itroduce the [R,L] sums ad defie the cyclic averages of the regular polygos ad the Platoic solids. We prove the mai property of the cyclic averages the equality of them for various regular polygos ad Platoic solids. By meas of the cyclic averages the distaces of a arbitrary poit to the vertices of the regular polygos the plae case ad the Platoic solids the space case are ivestigated. It is foud all cases of costat sum of like powers of the distaces, whe the locus is a circle a sphere, established the metrical relatios for regular polygos Platoic solids, which were kow i special cases oly. Ratioal distaces problem solved for the = 4 case. Ackowledgemet The author would like to thak Georgia Youg Scietists Uio GYSU. This research was fuded by GYSU. Cotract o

24 Refereces [] T. M. Apostol ad M. A. Matsakaia, Sums of squares of distaces. Math Horizos 9. 00,. [] T. M. Apostol ad M. A. Matsakaia, Sums of squares of distaces i m-space. America Mathematical Mothly 0 003, o. 6, [3] M. Garder, Mathematical circus. Mathematical Associatio of America, 99. [4] B. J. McCarti, Mysteries of the Equilateral Triagle. Hikari Ltd., Ruse, 00. [5] J. Beti, Regular polygoal distaces. The Mathematical Gazette 8 997, o. 49, [6] R. K. Guy, Usolved Problems i Number Theory. Spriger-Verlag, New York Berli, 98. [7] T. G. Berry, Poits at ratioal distace from the corers of a uit square. Aali della Scuola Normale Superiore di Pisa. Classe di Scieze. Serie IV 7 990, o. 4, [8] R. K. Guy, Tilig the square with ratioal triagles. Number Theory ad Applicatios, RA Molli ed., NATO Adv. Study Ist. Ser. C , [9] R. Barbara ad A. Karam, The ratioal distace problem for equilateral triagles. Commuicatios i Mathematics ad Applicatios 9 08, o., [0] R. Barbara, Poits at ratioal distace from the vertices of a uit polygo. Bulleti of the Iraia Mathematical Society , o., [] P. Tagsupphathawat, Algebraic trigoometric values at ratioal multipliers of π. Acta et Commetatioes Uiversitatis Tartuesis de Mathematica 8 04, o., 9 8. [] J. Beti, Regular simplicial distaces. The Mathematical Gazette , p. 06. [3] P.-S. Park, Regular polytopic distaces. Forum Geometricorum 6 06,

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